Quadratic equations

Start by writing $2x^{2} -7 x+3 = (x + \square )(x + \square )$. Choose any two factors of $\text{2}$ that can be the coefficients of $x$ and any two factors of $\text{3}$ that can replace the $\square$'s such that the two expressions on either side of the '$=$' sign become equal.

The correct answer is: $\left( x-3 \right) \left( 2x-1 \right)$.

Start by writing $6f^{2} - f-2 = (f + \square )(f + \square )$. Choose any two factors of $\text{6}$ that can be the coefficients of $f$ and any two factors of $-\text{2}$ that can replace the $\square$'s such that the two expressions on either side of the '$=$' sign become equal.

The correct answer is: $\left( 3f-2 \right) \left( 2f+1 \right)$.

Start by writing $3y^{2} +14 y-5 = (y + \square )(y + \square )$. Choose any two factors of $\text{3}$ that can be the coefficients of $y$ and any two factors of $-\text{5}$ that can replace the $\square$'s such that the two expressions on either side of the '$=$' sign become equal.

The correct answer is: $\left( y+5 \right) \left( 3y-1 \right)$.

This question is about factorisation: we have the equation $x^{2} - 4 x = x \left(\underline{\; ? \;} - 4\right)$ and we must fill in the missing value to make the equation true.

Factorisation is the inverse (the reverse) of the distributive law. When we factorise, we are "undoing" the distributive law. In other words, factorisation is when you "un-expand" or "undistribute" an expression. The picture below shows the relationship between distributing and factorising. When you multiply into the brackets you are distributing; when you put a number back outside of the brackets, you are factorising.

The equation given is: $x^{2} - 4 x = x \left(\underline{\; ? \;} - 4\right)$. We need the value which comes first in the brackets. Think about the distributive law: we need a number so that $x \, \times \, ?$ will be equal to $x^{2}$. This means that the missing value must be $x$.

Here are some important facts about factorisation:

• the number of terms in the brackets is equal to the number of terms in the original expression
• the value outside of the brackets is the HCF (Highest Common Factor) of the terms in $x^{2} - 4 x$
• we can check the answer with the distributive law:
  $$\begin{array}{cl} x \left(x - 4\right) & \quad \longleftarrow \; ^{\text{Expand the}}_{\text{brackets}} \\ x^{2} - 4 x & \quad \longleftarrow \; ^{\text{Cool - this is the}}_{\text{original expression}} \end{array}$$

The correct answer for the missing value is: $x$.

This question is about factorisation: we have the equation $3 x + 9 = 3 \left(x \, + \, \underline{\; ? \;} \right)$ and we must fill in the missing value to make the equation true.

Factorisation is the inverse (the reverse) of the distributive law. When we factorise, we are "undoing" the distributive law. In other words, factorisation is when you "un-expand" or "undistribute" an expression. The picture below shows the relationship between distributing and factorising. When you multiply into the brackets you are distributing; when you put a number back outside of the brackets, you are factorising.

The equation given is: $3 x + 9 = 3 \left(x \, + \, \underline{\; ? \;} \right)$. We need the value which comes second in the brackets. Think about the distributive law: we need a number so that $3 \, \times \, ?$ will be equal to $9$. This means that the missing value must be $3$.

Here are some important facts about factorisation:

• the number of terms in the brackets is equal to the number of terms in the original expression
• the value outside of the brackets is the HCF (Highest Common Factor) of the terms in $3 x + 9$
• we can check the answer with the distributive law:
  $$\begin{array}{cl} 3 \left(x + 3\right) & \quad \longleftarrow \; ^{\text{Expand the}}_{\text{brackets}} \\ 3 x + 9 & \quad \longleftarrow \; ^{\text{Cool - this is the}}_{\text{original expression}} \end{array}$$

The correct answer for the missing value is: $3$.

This question is about factorisation: we have the equation $x^{2} - 2 x = x \left(\underline{\; ? \;} - 2\right)$ and we must fill in the missing value to make the equation true.

Factorisation is the inverse (the reverse) of the distributive law. When we factorise, we are "undoing" the distributive law. In other words, factorisation is when you "un-expand" or "undistribute" an expression. The picture below shows the relationship between distributing and factorising. When you multiply into the brackets you are distributing; when you put a number back outside of the brackets, you are factorising.

The equation given is: $x^{2} - 2 x = x \left(\underline{\; ? \;} - 2\right)$. We need the value which comes first in the brackets. Think about the distributive law: we need a number so that $x \, \times \, ?$ will be equal to $x^{2}$. This means that the missing value must be $x$.

Here are some important facts about factorisation:

• the number of terms in the brackets is equal to the number of terms in the original expression
• the value outside of the brackets is the HCF (Highest Common Factor) of the terms in $x^{2} - 2 x$
• we can check the answer with the distributive law:
  $$\begin{array}{cl} x \left(x - 2\right) & \quad \longleftarrow \; ^{\text{Expand the}}_{\text{brackets}} \\ x^{2} - 2 x & \quad \longleftarrow \; ^{\text{Cool - this is the}}_{\text{original expression}} \end{array}$$

The correct answer for the missing value is: $x$.

To factorise this expression we must take out the highest common factor (HCF) from all the terms.

To factorise this expression we must take out the highest common factor (HCF) from all the terms. In this case the HCF is $3p$, so we get:

To factorise this expression we must take out the highest common factor (HCF) from all the terms.

To factorise this expression we must take out the highest common factor (HCF) from all the terms. In this case the HCF is $12p$, so we get:

To factorise this expression we must take out the highest common factor (HCF) from all the terms.

To factorise this expression we must take out the highest common factor (HCF) from all the terms. In this case the HCF is $7f$, so we get:

The question asks you to factorise the expression $5 x^{2} + x - 4$. Factorising an expression usually requires patience, but in this case it is worse than usual because the quadratic coefficient is not $1$: that adds to the complexity of the question.

Factorisation typically requires a trial and error approach, but you can usually use the quadratic term and the signs to start breaking down the factors. In this case:

• the $5 x^{2}$ term means that the factors must start with $5 x$ and $x$
• since the constant term is negative, the factors must have opposite signs (one positive and one negative)

That gives us two options at this point:

Note: these are two possible answer for the question. In the first option you have $5 x$ plus something while in the second you have $5 x$ minus something; when you do FOIL with these two situations, you will not get the same answer. In other words, one of them will turn out to be right and the other will turn out to be wrong. Right now we do not know which will be correct.

(If you are not sure where these values came from, imagine starting FOIL with these two brackets and compare to the expression $5 x^{2} + x - 4$.)

Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, they should have a product of $4$ so that we can get the last term in $5 x^{2} + x - 4$. (To be completely accurate, the numbers should have a product of $-4$, but we have already taken account of the negative in the signs in the binomials.)

For example, see what happens if you try the numbers $1$ and $4$. Remember that there are two options for where to put the signs in the binomials:

Shucks - neither result is equal to $5 x^{2} + x - 4$ (the $x$-terms are wrong in both options). That means that neither of the binomial options above is correct. Shame, you need to try another combination of numbers which have a product of $4$; and remember that there are two ways the signs can sit in the binomials.

The correct answer is: $\left(5 x - 4\right) \left(x + 1\right)$.

The question asks you to factorise the expression $2 x^{2} + 5 x - 3$. Factorising an expression usually requires patience, but in this case it is worse than usual because the quadratic coefficient is not $1$: that adds to the complexity of the question.

Factorisation typically requires a trial and error approach, but you can usually use the quadratic term and the signs to start breaking down the factors. In this case:

• the $2 x^{2}$ term means that the factors must start with $2 x$ and $x$
• since the constant term is negative, the factors must have opposite signs (one positive and one negative)

That gives us two options at this point:

Note: these are two possible answer for the question. In the first option you have $2 x$ plus something while in the second you have $2 x$ minus something; when you do FOIL with these two situations, you will not get the same answer. In other words, one of them will turn out to be right and the other will turn out to be wrong. Right now we do not know which will be correct.

(If you are not sure where these values came from, imagine starting FOIL with these two brackets and compare to the expression $2 x^{2} + 5 x - 3$.)

Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, they should have a product of $3$ so that we can get the last term in $2 x^{2} + 5 x - 3$. (To be completely accurate, the numbers should have a product of $-3$, but we have already taken account of the negative in the signs in the binomials.)

For example, see what happens if you try the numbers $1$ and $3$. Remember that there are two options for where to put the signs in the binomials:

Nice! The answer on the right is the trinomial you needed. That means the binomials on the right side must be correct! It is quite nice that it worked out on the first try: it often takes more than one try.

The correct answer is: $\left(2 x - 1\right) \left(x + 3\right)$.

The question asks you to factorise the expression $7 x^{2} - 4 x - 3$. Factorising an expression usually requires patience, but in this case it is worse than usual because the quadratic coefficient is not $1$: that adds to the complexity of the question.

Factorisation typically requires a trial and error approach, but you can usually use the quadratic term and the signs to start breaking down the factors. In this case:

• the $7 x^{2}$ term means that the factors must start with $7 x$ and $x$
• since the constant term is negative, the factors must have opposite signs (one positive and one negative)

That gives us two options at this point:

Note: these are two possible answer for the question. In the first option you have $7 x$ plus something while in the second you have $7 x$ minus something; when you do FOIL with these two situations, you will not get the same answer. In other words, one of them will turn out to be right and the other will turn out to be wrong. Right now we do not know which will be correct.

(If you are not sure where these values came from, imagine starting FOIL with these two brackets and compare to the expression $7 x^{2} - 4 x - 3$.)

Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, they should have a product of $3$ so that we can get the last term in $7 x^{2} - 4 x - 3$. (To be completely accurate, the numbers should have a product of $-3$, but we have already taken account of the negative in the signs in the binomials.)

For example, see what happens if you try the numbers $1$ and $3$. Remember that there are two options for where to put the signs in the binomials:

Shucks - neither result is equal to $7 x^{2} - 4 x - 3$ (the $x$-terms are wrong in both options). That means that neither of the binomial options above is correct. Shame, you need to try another combination of numbers which have a product of $3$; and remember that there are two ways the signs can sit in the binomials.

The correct answer is: $\left(7 x + 3\right) \left(x - 1\right)$.

Use the distributive law to expand the expressions in the table. The correct choice must be equal to $2 x^{2} + 3 x$ when it is expanded.

This question is about factorisation. We have the expression $2 x^{2} + 3 x$ and we need to select the choice in the table which is the correctly factorised version of that expression.

Factorisation is the inverse (the reverse) of the distributive law. Factorising means "undoing" the distributive law. In other words, factorisation is when you "un-expand" or "undistribute" an expression. The picture below shows the relationship between distributing and factorising. When we multiply into the brackets we are distributing; when we put a number back outside of the brackets, we are factorising.

We can use the relationship above to find the correct answer to this question! Just start by expanding the first choice in the table with the distributive law to see if it is equal to $2 x^{2} + 3 x$:

Shucks: it did not work out correctly. We needed the expression $2 x^{2} + 3 x$ but instead we got $6 x^{2} + 3 x$. Move on to checking the second choice in the table: expand it using the distributive law. Once you find the expression which is equal to $2 x^{2} + 3 x$ when you expand it, you have the correct answer.

In this question the correct answer is $x \left(2 x + 3\right)$ because when we distribute the $x$ into the brackets we get $2 x^{2} + 3 x$.

The correct choice from the table is: C.

Use the distributive law to expand the expressions in the table. The correct choice must be equal to $3 x^{2} + 2 x$ when it is expanded.

This question is about factorisation. We have the expression $3 x^{2} + 2 x$ and we need to select the choice in the table which is the correctly factorised version of that expression.

Factorisation is the inverse (the reverse) of the distributive law. Factorising means "undoing" the distributive law. In other words, factorisation is when you "un-expand" or "undistribute" an expression. The picture below shows the relationship between distributing and factorising. When we multiply into the brackets we are distributing; when we put a number back outside of the brackets, we are factorising.

We can use the relationship above to find the correct answer to this question! Just start by expanding the first choice in the table with the distributive law to see if it is equal to $3 x^{2} + 2 x$:

Fantastic! When we multiply out the first expression we find it is equal to $3 x^{2} + 2 x$; that means it is the correctly factorised version of the expression.

The correct choice from the table is: A.

Use the distributive law to expand the expressions in the table. The correct choice must be equal to $4 x^{2} + 3 x$ when it is expanded.

This question is about factorisation. We have the expression $4 x^{2} + 3 x$ and we need to select the choice in the table which is the correctly factorised version of that expression.

Factorisation is the inverse (the reverse) of the distributive law. Factorising means "undoing" the distributive law. In other words, factorisation is when you "un-expand" or "undistribute" an expression. The picture below shows the relationship between distributing and factorising. When we multiply into the brackets we are distributing; when we put a number back outside of the brackets, we are factorising.

We can use the relationship above to find the correct answer to this question! Just start by expanding the first choice in the table with the distributive law to see if it is equal to $4 x^{2} + 3 x$:

Shucks: it did not work out correctly. We needed the expression $4 x^{2} + 3 x$ but instead we got $4 x^{2} + 12 x$. Move on to checking the second choice in the table: expand it using the distributive law. Once you find the expression which is equal to $4 x^{2} + 3 x$ when you expand it, you have the correct answer.

In this question the correct answer is $x \left(4 x + 3\right)$ because when we distribute the $x$ into the brackets we get $4 x^{2} + 3 x$.

The correct choice from the table is: C.

You have the expression $12 x^{2} y - 24 y^{2} - 20 y$ and you must factorise it. Factorisation is the inverse of the distributive law, which means it is when you "un-expand" or "undistribute" an expression.

The first thing you need to do is find the HCF of the expression. However, the expression is a trinomial, so you must find the HCF of all three terms! The terms are $12 x^{2} y$, $- 24 y^{2}$ and $- 20 y$. The HCF for these three terms is $4 y$.

Once you know the HCF, you can factorise the expression. Start by putting the HCF in front of a pair of brackets, like this: $12 x^{2} y - 24 y^{2} - 20 y = 4 y \times \left( \textrm{more stuff} \right)$. If you take $4 y$ out of each term in the trinomial you will get $4 y \left(3 x^{2} - 6 y - 5\right)$.

Remember that you should always check your answer when you factorise. To check the answer, distribute the HCF back into the brackets to make sure that everything agrees with the original expression. Remember, you must distribute the $4 y$ to all three terms in the brackets.

The correct answer is: $4 y \left(3 x^{2} - 6 y - 5\right)$.

You have the expression $5 a^{2} b - 3 b c + 4$ and you must factorise it. Factorisation is the inverse of the distributive law, which means it is when you "un-expand" or "undistribute" an expression.

The first thing you need to do is find the HCF of the expression. However, the expression is a trinomial, so you must find the HCF of all three terms! The terms are $5 a^{2} b$, $- 3 b c$ and $4$. These three terms have no factors which are shared in all three of the terms; therefore, the HCF is $1$.

Once you know the HCF, you can factorise the expression. However, this is a special situation: the HCF is $1$. That means that there are no common factors in the terms of the expression. Therefore, the expression must be prime!

A prime quantity has only two factors: $1$ and itself. This expression has no other factors, only the HCF of $1$ and the expression itself.

The correct answer is: prime.

You have the expression $9 x^{2} + 18 x + 15$ and you must factorise it. Factorisation is the inverse of the distributive law, which means it is when you "un-expand" or "undistribute" an expression.

The first thing you need to do is find the HCF of the expression. However, the expression is a trinomial, so you must find the HCF of all three terms! The terms are $9 x^{2}$, $18 x$ and $15$. The HCF for these three terms is $3$.

Once you know the HCF, you can factorise the expression. Start by putting the HCF in front of a pair of brackets, like this: $9 x^{2} + 18 x + 15 = 3 \times \left( \textrm{more stuff} \right)$. If you take $3$ out of each term in the trinomial you will get $3 \left(3 x^{2} + 6 x + 5\right)$.

Remember that you should always check your answer when you factorise. To check the answer, distribute the HCF back into the brackets to make sure that everything agrees with the original expression. Remember, you must distribute the $3$ to all three terms in the brackets.

The correct answer is: $3 \left(3 x^{2} + 6 x + 5\right)$.

We have the expression $x^{2} + 7 x + 10$ and we must factorise it.

To factorise an expression means to write the expression as a product. For example, if you factorise the number $10$, you would write $2 \cdot 5$... $2$ and $5$ are factors of $10$. In the case of a trinomial like $x^{2} + 7 x + 10$ factorising usually leads to two binomials: we want two binomials which have a product equal to $x^{2} + 7 x + 10$, just like $2$ and $5$ have a product of $10$.

To work out the binomial factors requires patience, because there is no single calculation which will give us the answer. Rather we must take the time to work out the answer by trial and error (guess & check).

The first step is to set up the answer based on the $x^{2}$ term in the trinomial. It tells us that the binomials must be like this:

Imagine starting FOIL with these two brackets: you will get $x^{2}$ from the two $x$'s, which is the first part of $x^{2} + 7 x + 10$.

So far we have this: $x^{2} + 7 x + 10 = (x + \underline{\quad})(x + \underline{\quad})$. Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, we must check the binomials with FOIL to make sure that everything works out correctly.

We should try numbers which have a product of $10$. This is because of the "Lasts" term in FOIL. For example, let's check what happens if we try the numbers $1$ and $10$:

Shucks - we did not get the trinomial we needed because the $x$ term is not correct. That means that the binomials $\left(x + 1\right)$ and $\left(x + 10\right)$ cannot be the correct factors.

This is where patience is required: we need to try two other numbers which have a product of $10$ and see if those numbers work out to make the trinomial we need. We must continue this process until we find two binomials which agree with the trinomial. The correct pair of binomials is $\left(x + 5\right)\left(x + 2\right)$ because $\left(x + 5\right)\left(x + 2\right) = x^{2} + 7 x + 10$.

The correct answer is: $\left(x + 5\right) \left(x + 2\right)$.

We have the expression $x^{2} + 9 x + 18$ and we must factorise it.

To factorise an expression means to write the expression as a product. For example, if you factorise the number $10$, you would write $2 \cdot 5$... $2$ and $5$ are factors of $10$. In the case of a trinomial like $x^{2} + 9 x + 18$ factorising usually leads to two binomials: we want two binomials which have a product equal to $x^{2} + 9 x + 18$, just like $2$ and $5$ have a product of $10$.

To work out the binomial factors requires patience, because there is no single calculation which will give us the answer. Rather we must take the time to work out the answer by trial and error (guess & check).

The first step is to set up the answer based on the $x^{2}$ term in the trinomial. It tells us that the binomials must be like this:

Imagine starting FOIL with these two brackets: you will get $x^{2}$ from the two $x$'s, which is the first part of $x^{2} + 9 x + 18$.

So far we have this: $x^{2} + 9 x + 18 = (x + \underline{\quad})(x + \underline{\quad})$. Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, we must check the binomials with FOIL to make sure that everything works out correctly.

We should try numbers which have a product of $18$. This is because of the "Lasts" term in FOIL. For example, let's check what happens if we try the numbers $3$ and $6$:

Nice! We got the trinomial we needed, so the binomials must be correct!

The correct answer is: $\left(x + 3\right) \left(x + 6\right)$.

We have the expression $x^{2} + 8 x + 15$ and we must factorise it.

To factorise an expression means to write the expression as a product. For example, if you factorise the number $10$, you would write $2 \cdot 5$... $2$ and $5$ are factors of $10$. In the case of a trinomial like $x^{2} + 8 x + 15$ factorising usually leads to two binomials: we want two binomials which have a product equal to $x^{2} + 8 x + 15$, just like $2$ and $5$ have a product of $10$.

To work out the binomial factors requires patience, because there is no single calculation which will give us the answer. Rather we must take the time to work out the answer by trial and error (guess & check).

The first step is to set up the answer based on the $x^{2}$ term in the trinomial. It tells us that the binomials must be like this:

Imagine starting FOIL with these two brackets: you will get $x^{2}$ from the two $x$'s, which is the first part of $x^{2} + 8 x + 15$.

So far we have this: $x^{2} + 8 x + 15 = (x + \underline{\quad})(x + \underline{\quad})$. Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, we must check the binomials with FOIL to make sure that everything works out correctly.

We should try numbers which have a product of $15$. This is because of the "Lasts" term in FOIL. For example, let's check what happens if we try the numbers $1$ and $15$:

Shucks - we did not get the trinomial we needed because the $x$ term is not correct. That means that the binomials $\left(x + 1\right)$ and $\left(x + 15\right)$ cannot be the correct factors.

This is where patience is required: we need to try two other numbers which have a product of $15$ and see if those numbers work out to make the trinomial we need. We must continue this process until we find two binomials which agree with the trinomial. The correct pair of binomials is $\left(x + 3\right)\left(x + 5\right)$ because $\left(x + 3\right)\left(x + 5\right) = x^{2} + 8 x + 15$.

The correct answer is: $\left(x + 3\right) \left(x + 5\right)$.

Start by writing $h^{2}-9h\text{+8} = (h + \square )(h + \square )$. Choose any two factors of $\text{+8}$ that can replace the $\square$'s such that the two expressions on either side of the '$=$' sign become equal.

Factorisation usually requires patience: it is not something you can do by following a formula. So give yourself time to find the answer if it does not come quickly.

The most useful thing to look at to start factorisation is the constant term; in this case it is equal to $\text{+8}$. That means that when you factorise you must find two numbers which have a product of ('product' means multiply) $+8$. The bad news is that those two numbers must also agree with the other terms in the question, which often makes it a bit complex.

In this case it works out like this:

Start by writing $s^{2}+3s-\text{70} = (s + \square )(s + \square )$. Choose any two factors of $-\text{70}$ that can replace the $\square$'s such that the two expressions on either side of the '$=$' sign become equal.

Factorisation usually requires patience: it is not something you can do by following a formula. So give yourself time to find the answer if it does not come quickly.

The most useful thing to look at to start factorisation is the constant term; in this case it is equal to $-\text{70}$. That means that when you factorise you must find two numbers which have a product of ('product' means multiply) $-70$. The bad news is that those two numbers must also agree with the other terms in the question, which often makes it a bit complex.

In this case it works out like this:

Start by writing $v^{2}-12v\text{+32} = (v + \square )(v + \square )$. Choose any two factors of $\text{+32}$ that can replace the $\square$'s such that the two expressions on either side of the '$=$' sign become equal.

Factorisation usually requires patience: it is not something you can do by following a formula. So give yourself time to find the answer if it does not come quickly.

The most useful thing to look at to start factorisation is the constant term; in this case it is equal to $\text{+32}$. That means that when you factorise you must find two numbers which have a product of ('product' means multiply) $+32$. The bad news is that those two numbers must also agree with the other terms in the question, which often makes it a bit complex.

In this case it works out like this:

1. use the signs in the trinomial to figure out what the signs in the binomials should be
2. be careful to conclude that the trinomial is prime if you try a few combinations and they all fail: you need to try all possible combinations of numbers to get the quadratic and the constant term. If none of those combinations work correctly, then the trinomial is prime

This question is tough: both the quadratic coefficient and the constant term are composite numbers (not prime), which means there are a lot of combinations of numbers that you may need to try to get the answer. Ugh.

Use the signs in the trinomial as a guide for the signs in the binomials: since the constant term is negative, the signs in the binomials must be opposites (one positive and one negative). Once you know what to expect with the signs, start working with the numbers to try to find the factors.

Based on the trinomial, here is one possible starting point:

If this option does not lead to the correct factors, you may have to come back to this starting point and make a change. It is possible that the trinomial is prime, but remember that you must try all possible combinations of numbers to get the quadratic and the constant term from the trinomial; if none of those combinations work correctly, then the trinomial is prime.

The correct factors are $\left(3 z - 4\right) \left(5 z + 3\right)$, because when you expand the brackets it works out:

The correct answer is: $\left(3 z - 4\right) \left(5 z + 3\right)$.

1. use the signs in the trinomial to figure out what the signs in the binomials should be
2. be careful to conclude that the trinomial is prime if you try a few combinations and they all fail: you need to try all possible combinations of numbers to get the quadratic and the constant term. If none of those combinations work correctly, then the trinomial is prime

This question is tough: both the quadratic coefficient and the constant term are composite numbers (not prime), which means there are a lot of combinations of numbers that you may need to try to get the answer. Ugh.

Use the signs in the trinomial as a guide for the signs in the binomials: since all of the terms in the trinomial are positive, the factors must both have plus signs. Once you know what to expect with the signs, start working with the numbers to try to find the factors.

Based on the trinomial, here is one possible starting point:

If this option does not lead to the correct factors, you may have to come back to this starting point and make a change. It is possible that the trinomial is prime, but remember that you must try all possible combinations of numbers to get the quadratic and the constant term from the trinomial; if none of those combinations work correctly, then the trinomial is prime.

The correct factors are $\left(5 x + 2\right) \left(3 x + 8\right)$, because when you expand the brackets it works out:

The correct answer is: $\left(5 x + 2\right) \left(3 x + 8\right)$.

1. use the signs in the trinomial to figure out what the signs in the binomials should be
2. be careful to conclude that the trinomial is prime if you try a few combinations and they all fail: you need to try all possible combinations of numbers to get the quadratic and the constant term. If none of those combinations work correctly, then the trinomial is prime

This question is tough: both the quadratic coefficient and the constant term are composite numbers (not prime), which means there are a lot of combinations of numbers that you may need to try to get the answer. Ugh.

Use the signs in the trinomial as a guide for the signs in the binomials: since all of the terms in the trinomial are positive, the factors must both have plus signs. Once you know what to expect with the signs, start working with the numbers to try to find the factors.

Based on the trinomial, here is one possible starting point:

If this option does not lead to the correct factors, you may have to come back to this starting point and make a change. It is possible that the trinomial is prime, but remember that you must try all possible combinations of numbers to get the quadratic and the constant term from the trinomial; if none of those combinations work correctly, then the trinomial is prime.

The correct factors are $\left(4 x + 7\right) \left(5 x + 2\right)$, because when you expand the brackets it works out:

The correct answer is: $\left(4 x + 7\right) \left(5 x + 2\right)$.

You have the expression $x^{2} + x - 6$ and you must factorise it. To work out the binomial factors often requires patience: there is no single calculation which will give you the answer. Rather you must take the time to work out the answer by trial and error (guess & check).

For the expression $x^{2} + x - 6$, the first and last terms both give big hints about the binomials that you need. The first term suggests that the factors must both have $x$ in them: $\left( x \qquad \right)\left( x \qquad \right)$.

The last term in $x^{2} + x - 6$ tells you about the signs in the binomials. Remember that the number $-6$ usually comes from multiplying the numbers in the binomials (when you do FOIL). Therefore, the binomials must contain one negative sign and one positive sign. Therefore the binomials must be:

So at this point we have $x^{2} + x - 6 = (x + \underline{\quad})(x - \underline{\quad})$.

Now you need to figure out which numbers belong in the empty spaces. Whatever numbers you try to put in the binomials, you must check the binomials with FOIL to make sure that everything works out correctly.

You should try numbers which have a product of $-6$. For example, let's check what happens with the numbers $1$ and $-6$:

Nope! This is not the trinomial you needed: the $x$-term is not correct. (The other two terms are correct, but maths is very strict - you must get the trinomial $x^{2} + x - 6$ exactly, not almost the same trinomial.) That means that the binomials $\left(x + 1\right)$ and $\left(x - 6\right)$ cannot be the correct factors.

Now stay patient: you need to try two other numbers which have a product of $-6$ and see if those numbers work out to make the trinomial $x^{2} + x - 6$. You must continue this process until you find two binomials which agree with the trinomial.

The correct answer is: $\left(x + 3\right) \left(x - 2\right)$. You can also write the answer as $\left(x - 2\right) \left(x + 3\right)$.

You have the expression $x^{2} - 4 x - 21$ and you must factorise it. To work out the binomial factors often requires patience: there is no single calculation which will give you the answer. Rather you must take the time to work out the answer by trial and error (guess & check).

For the expression $x^{2} - 4 x - 21$, the first and last terms both give big hints about the binomials that you need. The first term suggests that the factors must both have $x$ in them: $\left( x \qquad \right)\left( x \qquad \right)$.

The last term in $x^{2} - 4 x - 21$ tells you about the signs in the binomials. Remember that the number $-21$ usually comes from multiplying the numbers in the binomials (when you do FOIL). Therefore, the binomials must contain one negative sign and one positive sign. Therefore the binomials must be:

So at this point we have $x^{2} - 4 x - 21 = (x + \underline{\quad})(x - \underline{\quad})$.

Now you need to figure out which numbers belong in the empty spaces. Whatever numbers you try to put in the binomials, you must check the binomials with FOIL to make sure that everything works out correctly.

You should try numbers which have a product of $-21$. For example, let's check what happens with the numbers $1$ and $-21$:

Nope! This is not the trinomial you needed: the $x$-term is not correct. (The other two terms are correct, but maths is very strict - you must get the trinomial $x^{2} - 4 x - 21$ exactly, not almost the same trinomial.) That means that the binomials $\left(x + 1\right)$ and $\left(x - 21\right)$ cannot be the correct factors.

Now stay patient: you need to try two other numbers which have a product of $-21$ and see if those numbers work out to make the trinomial $x^{2} - 4 x - 21$. You must continue this process until you find two binomials which agree with the trinomial.

The correct answer is: $\left(x - 7\right) \left(x + 3\right)$. You can also write the answer as $\left(x + 3\right) \left(x - 7\right)$.

You have the expression $x^{2} - 5 x - 14$ and you must factorise it. To work out the binomial factors often requires patience: there is no single calculation which will give you the answer. Rather you must take the time to work out the answer by trial and error (guess & check).

For the expression $x^{2} - 5 x - 14$, the first and last terms both give big hints about the binomials that you need. The first term suggests that the factors must both have $x$ in them: $\left( x \qquad \right)\left( x \qquad \right)$.

The last term in $x^{2} - 5 x - 14$ tells you about the signs in the binomials. Remember that the number $-14$ usually comes from multiplying the numbers in the binomials (when you do FOIL). Therefore, the binomials must contain one negative sign and one positive sign. Therefore the binomials must be:

So at this point we have $x^{2} - 5 x - 14 = (x + \underline{\quad})(x - \underline{\quad})$.

Now you need to figure out which numbers belong in the empty spaces. Whatever numbers you try to put in the binomials, you must check the binomials with FOIL to make sure that everything works out correctly.

You should try numbers which have a product of $-14$. For example, let's check what happens with the numbers $1$ and $-14$:

Nope! This is not the trinomial you needed: the $x$-term is not correct. (The other two terms are correct, but maths is very strict - you must get the trinomial $x^{2} - 5 x - 14$ exactly, not almost the same trinomial.) That means that the binomials $\left(x + 1\right)$ and $\left(x - 14\right)$ cannot be the correct factors.

Now stay patient: you need to try two other numbers which have a product of $-14$ and see if those numbers work out to make the trinomial $x^{2} - 5 x - 14$. You must continue this process until you find two binomials which agree with the trinomial.

The correct answer is: $\left(x - 7\right) \left(x + 2\right)$. You can also write the answer as $\left(x + 2\right) \left(x - 7\right)$.

We have the expression $3 x - 15$ and we must factorise it. Factorisation means that we need to "undo" the distributive law - factorisation is the inverse of distributing. In other words, factorisation is when you "un-expand" or "undistribute" an expression.

The first thing we need to do is find the HCF of the two terms in the expression. (Remember that the HCF is the largest number or value which fits into both of the terms.) For the expression $3 x - 15$ the HCF of the terms is $3$.

Once you know the HCF, you can factorise the expression. As mentioned above, factorisation is when we "undo" the distributive law. The picture below shows the relationship between distributing and factorising. When you multiply into the brackets you are distributing; when you put a number back outside of the brackets, you are factorising.

In this question we are factorising, so start by putting the HCF in front of a pair of brackets, like this: $3 x - 15 = 3 \left(\text{more stuff}\right)$. Then you must fill in the brackets so that the expressions are equal.

In this case, if you take $3$ out of the terms in the expression you will get $3 \left(x - 5\right)$. Can you see why factorisation means "undistributing"?

Checking the answer: When we factorise, we should always take the time necessary to check the answer. To check the answer we need to distribute the HCF, $3$, back into the brackets to make sure that everything agrees with the original expression.

The correct answer is: $3 \left(x - 5\right)$.

We have the expression $9 x + 9 y$ and we must factorise it. Factorisation means that we need to "undo" the distributive law - factorisation is the inverse of distributing. In other words, factorisation is when you "un-expand" or "undistribute" an expression.

The first thing we need to do is find the HCF of the two terms in the expression. (Remember that the HCF is the largest number or value which fits into both of the terms.) For the expression $9 x + 9 y$ the HCF of the terms is $9$.

Once you know the HCF, you can factorise the expression. As mentioned above, factorisation is when we "undo" the distributive law. The picture below shows the relationship between distributing and factorising. When you multiply into the brackets you are distributing; when you put a number back outside of the brackets, you are factorising.

In this question we are factorising, so start by putting the HCF in front of a pair of brackets, like this: $9 x + 9 y = 9 \left(\text{more stuff}\right)$. Then you must fill in the brackets so that the expressions are equal.

In this case, if you take $9$ out of the terms in the expression you will get $9 \left(x + y\right)$. Can you see why factorisation means "undistributing"?

Checking the answer: When we factorise, we should always take the time necessary to check the answer. To check the answer we need to distribute the HCF, $9$, back into the brackets to make sure that everything agrees with the original expression.

The correct answer is: $9 \left(x + y\right)$.

We have the expression $2 x^{2} + 3 x$ and we must factorise it. Factorisation means that we need to "undo" the distributive law - factorisation is the inverse of distributing. In other words, factorisation is when you "un-expand" or "undistribute" an expression.

The first thing we need to do is find the HCF of the two terms in the expression. (Remember that the HCF is the largest number or value which fits into both of the terms.) For the expression $2 x^{2} + 3 x$ the HCF of the terms is $x$.

Once you know the HCF, you can factorise the expression. As mentioned above, factorisation is when we "undo" the distributive law. The picture below shows the relationship between distributing and factorising. When you multiply into the brackets you are distributing; when you put a number back outside of the brackets, you are factorising.

In this question we are factorising, so start by putting the HCF in front of a pair of brackets, like this: $2 x^{2} + 3 x = x \left(\text{more stuff}\right)$. Then you must fill in the brackets so that the expressions are equal.

In this case, if you take $x$ out of the terms in the expression you will get $x \left(2 x + 3\right)$. Can you see why factorisation means "undistributing"?

Checking the answer: When we factorise, we should always take the time necessary to check the answer. To check the answer we need to distribute the HCF, $x$, back into the brackets to make sure that everything agrees with the original expression.

The correct answer is: $x \left(2 x + 3\right)$.

You have the expression $6 a c + 9 b c$ and you must factorise it. Factorisation is the inverse of the distributive law, which means it is when you "un-expand" or "undistribute" an expression.

The first thing you need to do is find the HCF of the two terms in the expression. The two terms are $6 a c$ and $9 b c$. The HCF of the terms is $3 c$.

Once you know the HCF, you can factorise the expression. Start by putting the HCF in front of a pair of brackets, like this: $6 a c + 9 b c = 3 c \times \left( \textrm{more stuff} \right)$. If you take $3 c$ out of the terms in the expression you will get $3 c \left(2 a + 3 b\right)$.

Remember that you should always check your answer when you factorise. To check the answer, distribute the HCF back into the brackets to make sure that everything agrees with the original expression.

The correct answer is: $3 c \left(2 a + 3 b\right)$.

You have the expression $5 a^{4} - 4$ and you must factorise it. Factorisation is the inverse of the distributive law, which means it is when you "un-expand" or "undistribute" an expression.

The first thing you need to do is find the HCF of the two terms in the expression. The two terms are $5 a^{4}$ and $-4$. These two terms have nothing in common at all, so the HCF is $1$.

Once you know the HCF, you can factorise the expression. However, this is a special situation: the HCF is $1$. That means that there are no common factors in the terms of the expression. Therefore, the expression must be prime!

A prime quantity has only two factors: $1$ and itself. This expression has no other factors, only the HCF of $1$ and the expression itself.

The correct answer is: prime.

You have the expression $6 x^{2} z - 10 x y z$ and you must factorise it. Factorisation is the inverse of the distributive law, which means it is when you "un-expand" or "undistribute" an expression.

The first thing you need to do is find the HCF of the two terms in the expression. The two terms are $6 x^{2} z$ and $- 10 x y z$. The HCF of the terms is $2 x z$.

Once you know the HCF, you can factorise the expression. Start by putting the HCF in front of a pair of brackets, like this: $6 x^{2} z - 10 x y z = 2 x z \times \left( \textrm{more stuff} \right)$. If you take $2 x z$ out of the terms in the expression you will get $2 x z \left(3 x - 5 y\right)$.

Remember that you should always check your answer when you factorise. To check the answer, distribute the HCF back into the brackets to make sure that everything agrees with the original expression.

The correct answer is: $2 x z \left(3 x - 5 y\right)$.

You need to factorise the expression $4 x^{2} - 9$. To work out the binomial factors often requires patience: however, there is a pattern in this expression which lets you get straight to the answer!

The expression $4 x^{2} - 9$ is a difference of two squares. (Both terms are squared values, and the subtraction is a difference.) The difference of two squares pattern is linked directly to binomials which are identical except for the signs in the binomials:

The correct answer is: $\left(2 x + 3\right) \left(2 x - 3\right)$.

You need to factorise the expression $36 x^{2} - 25$. To work out the binomial factors often requires patience: however, there is a pattern in this expression which lets you get straight to the answer!

The expression $36 x^{2} - 25$ is a difference of two squares. (Both terms are squared values, and the subtraction is a difference.) The difference of two squares pattern is linked directly to binomials which are identical except for the signs in the binomials:

The correct answer is: $\left(6 x + 5\right) \left(6 x - 5\right)$.

You need to factorise the expression $25 x^{2} - 36$. To work out the binomial factors often requires patience: however, there is a pattern in this expression which lets you get straight to the answer!

The expression $25 x^{2} - 36$ is a difference of two squares. (Both terms are squared values, and the subtraction is a difference.) The difference of two squares pattern is linked directly to binomials which are identical except for the signs in the binomials:

The correct answer is: $\left(5 x + 6\right) \left(5 x - 6\right)$.

You must factorise the expression $10 x z - 4 y z - 12 z$. Start by figuring out the HCF of the expression. When dealing with a trinomial, the HCF must come from all three of the terms in the expression.

The HCF for these terms is $2 z$.

Now factorise the HCF out of the expression. If you take $2 z$ out of each term in the expression you will get $2 z \left(5 x - 2 y - 6\right)$.

Finally, check your answer. Do this by distributing the HCF back into the brackets to make sure you get the original expression. Remember, you must distribute the $2 z$ to each of the terms in the brackets.

The correct answer is: $2 z \left(5 x - 2 y - 6\right)$.

You must factorise the expression $12 x^{3} + 15 x^{2} y + 18 x^{2} z$. Start by figuring out the HCF of the expression. When dealing with a trinomial, the HCF must come from all three of the terms in the expression.

The HCF for these terms is $3 x^{2}$.

Now factorise the HCF out of the expression. If you take $3 x^{2}$ out of each term in the expression you will get $3 x^{2} \left(4 x + 5 y + 6 z\right)$.

Finally, check your answer. Do this by distributing the HCF back into the brackets to make sure you get the original expression. Remember, you must distribute the $3 x^{2}$ to each of the terms in the brackets.

The correct answer is: $3 x^{2} \left(4 x + 5 y + 6 z\right)$.

You must factorise the expression $- 15 x y z - 12 y^{2} z - 6 y z$. Start by figuring out the HCF of the expression. When dealing with a trinomial, the HCF must come from all three of the terms in the expression.

When all of the terms are negative, you should include the negative sign in the HCF. The HCF of the terms is $- 3 y z$.

Now factorise the HCF out of the expression. If you take $- 3 y z$ out of each term in the expression you will get $- 3 y z \left(5 x + 4 y + 2\right)$.

Finally, check your answer. Do this by distributing the HCF back into the brackets to make sure you get the original expression. Remember, you must distribute the $- 3 y z$ to each of the terms in the brackets.

The correct answer is: $- 3 y z \left(5 x + 4 y + 2\right)$.

Take out the common factor which is the expression in brackets.

We take out the common bracket from the two terms to get:

The order of the brackets does not matter, since $a \times b$ is the same as $b \times a$. Therefore the following is also correct:

Take out the common factor which is the expression in brackets.

We take out the common bracket from the two terms to get:

The order of the brackets does not matter, since $a \times b$ is the same as $b \times a$. Therefore the following is also correct:

Take out the common factor which is the expression in brackets.

We take out the common bracket from the two terms to get:

The order of the brackets does not matter, since $a \times b$ is the same as $b \times a$. Therefore the following is also correct:

1. Look at the constant term in the trinomial: it is negative, so the binomials must have opposite signs, like this: $\small \left(\underline{\quad} + \underline{\quad} \right) \left(\underline{\quad} - \underline{\quad} \right)$
2. It is possible that the trinomial is prime; you need to try all possible combinations of numbers to get the quadratic and the constant term; if none of those combinations work correctly, then the trinomial is prime

You have the expression $3 x^{2} - 4 x - 7$ and you must factorise it. This question also has the challenging possibility that the expression is prime. To determine if the expression is prime, you must try all possible combinations of numbers to get the quadratic and the constant term from the trinomial; if none of those combinations work correctly, then the trinomial is prime.

For the factors of $3 x^{2} - 4 x - 7$, the following things must be true:

• the $3 x^{2}$ term means that the factors must start with $3 x$ and $x$
• since the constant term is negative, the signs in the binomials must be opposites (one positive and one negative)

From that information, here is an intelligent starting point:

Remember that you may also need to try the option above with the signs swapped, $(3 x - \underline{\quad})(x + \underline{\quad})$.

At this point you must try some numbers in the binomials; the numbers should have a product of $7$. For example, see what happens if you try the numbers $1$ and $7$.

Unfortunately, those binomials are not the correct factors. If there are any other possible combinations of numbers to put into the binomials, you must try them to see which combination works out. For example, you can try swapping the signs in the binomials: $\left(3 x - 1\right) \left(x + 7\right)$ (which will lead to a different trinomial).

In the end, there is a way to factorise the trinomial, so it is not prime.

The correct answer is: $\left(3 x - 7\right) \left(x + 1\right)$.

1. The constant term is positive, but the linear term is negative. That means that the binomials must both have negative signs: $\small \left(\underline{\quad} - \underline{\quad} \right) \left(\underline{\quad} - \underline{\quad} \right)$
2. It is possible that the trinomial is prime; you need to try all possible combinations of numbers to get the quadratic and the constant term; if none of those combinations work correctly, then the trinomial is prime

You have the expression $3 x^{2} - 19 x + 10$ and you must factorise it. This question also has the challenging possibility that the expression is prime. To determine if the expression is prime, you must try all possible combinations of numbers to get the quadratic and the constant term from the trinomial; if none of those combinations work correctly, then the trinomial is prime.

For the factors of $3 x^{2} - 19 x + 10$, the following things must be true:

• the $3 x^{2}$ term means that the factors must start with $3 x$ and $x$
• since the constant term is positive but the linear term is negative, the signs in the binomials must both be negatives

From that information, here is an intelligent starting point:

At this point you must try some numbers in the binomials; the numbers should have a product of $10$. For example, see what happens if you try the numbers $10$ and $1$.

This is not the trinomial you needed because the $x$-term is not correct. That means that the binomials $\left(3 x - 10\right)$ and $\left(x - 1\right)$ cannot be the correct factors.

Now you must try another combination of numbers in the binomials. For example, you can try swapping the numbers in the binomials: $\left(3 x - 1\right) \left(x - 10\right)$. Remember that this will result in a different trinomial because there are different coefficients for $x$ in the binomials. In the end, though, there are no combinations of numbers which agree with everything in the trinomial. This expressoin is prime.

The correct answer is: prime.

1. The constant term is positive, but the linear term is negative. That means that the binomials must both have negative signs: $\small \left(\underline{\quad} - \underline{\quad} \right) \left(\underline{\quad} - \underline{\quad} \right)$
2. It is possible that the trinomial is prime; you need to try all possible combinations of numbers to get the quadratic and the constant term; if none of those combinations work correctly, then the trinomial is prime

You have the expression $5 x^{2} - 14 x + 9$ and you must factorise it. This question also has the challenging possibility that the expression is prime. To determine if the expression is prime, you must try all possible combinations of numbers to get the quadratic and the constant term from the trinomial; if none of those combinations work correctly, then the trinomial is prime.

For the factors of $5 x^{2} - 14 x + 9$, the following things must be true:

• the $5 x^{2}$ term means that the factors must start with $5 x$ and $x$
• since the constant term is positive but the linear term is negative, the signs in the binomials must both be negatives

From that information, here is an intelligent starting point:

At this point you must try some numbers in the binomials; the numbers should have a product of $9$. For example, see what happens if you try the numbers $9$ and $1$.

Great! We got the trinomial we needed, so the binomials must be correct.

The correct answer is: $\left(5 x - 9\right) \left(x - 1\right)$.

Start by identifying the highest common factor (HCF) of both terms.

We take out the highest common factor (HCF) from both terms. In this case, the HCF is $2a$.

When you have factorised the HCF out of the expression, there cannot be any more common factors remaining in the brackets!

Start by identifying the highest common factor (HCF) of both terms.

We take out the highest common factor (HCF) from both terms. In this case, the HCF is $2g$.

When you have factorised the HCF out of the expression, there cannot be any more common factors remaining in the brackets!

Start by identifying the highest common factor (HCF) of both terms.

We take out the highest common factor (HCF) from both terms. In this case, the HCF is $2h$.

When you have factorised the HCF out of the expression, there cannot be any more common factors remaining in the brackets!

You need to factorise the expression $25 a^{2} - 36$. Notice that all of the values in the expression are squared values (either square numbers or the square of a variable). Furthermore, the expression shows a difference (subtraction). All those facts together mean that the expression is a difference of two squares.

For the difference of two squares pattern, the binomials are always identical except for the signs in the binomials:

As always, you should check the binomials to make sure you get the exact expression you need.

The correct answer is: $\left(5 a + 6\right) \left(5 a - 6\right)$. The order of the binomials is not important, so you can also write the answer as $\left(5 a - 6\right) \left(5 a + 6\right)$.

You need to factorise the expression $49 - 16 y^{2}$. Notice that all of the values in the expression are squared values (either square numbers or the square of a variable). Furthermore, the expression shows a difference (subtraction). All those facts together mean that the expression is a difference of two squares.

For the difference of two squares pattern, the binomials are always identical except for the signs in the binomials:

As always, you should check the binomials to make sure you get the exact expression you need.

The correct answer is: $\left(7 + 4 y\right) \left(7 - 4 y\right)$. The order of the binomials is not important, so you can also write the answer as $\left(7 - 4 y\right) \left(7 + 4 y\right)$.

You need to factorise the expression $64 - 121 h^{2}$. Notice that all of the values in the expression are squared values (either square numbers or the square of a variable). Furthermore, the expression shows a difference (subtraction). All those facts together mean that the expression is a difference of two squares.

For the difference of two squares pattern, the binomials are always identical except for the signs in the binomials:

As always, you should check the binomials to make sure you get the exact expression you need.

The correct answer is: $\left(8 + 11 h\right) \left(8 - 11 h\right)$. The order of the binomials is not important, so you can also write the answer as $\left(8 - 11 h\right) \left(8 + 11 h\right)$.

Factorisation is the inverse of expanding something. When we factorise, we must break something apart into a product. For example, if you factorise the number $10$, you write $2 \cdot 5$. For a trinomial, it is common that the factors are binomials, like what is shown in this question.

To see if the binomials are the correct factors, we must expand the product to find out if they are equal to the trinomial $x^{2} + 4 x + 4$ or not. That means using FOIL:

We were supposed to get the trinomial $x^{2} + 4 x + 4$! However, the constant term and the $x$-term are both wrong. Therefore $x^{2} + 4 x + 4$ is not equal to $\left(x + 2\right)\left(x + 4\right)$

Notice that we checked the answer by expanding the binomials with FOIL and comparing the result to the trinomial in the question. Checking the answer with FOIL is a normal part of factorisation: you will do it over and over again when you are solving factorisation problems.

The correct choice is: False.

Factorisation is the inverse of expanding something. When we factorise, we must break something apart into a product. For example, if you factorise the number $10$, you write $2 \cdot 5$. For a trinomial, it is common that the factors are binomials, like what is shown in this question.

To see if the binomials are the correct factors, we must expand the product to find out if they are equal to the trinomial $x^{2} + 4 x + 4$ or not. That means using FOIL:

It works! The brackets multiply out to make the trinomial in the question. The brackets are the correct factors because $x^{2} + 4 x + 4$ is equal to $\left(x + 2\right)\left(x + 2\right)$.

Notice that we checked the answer by expanding the binomials with FOIL and comparing the result to the trinomial in the question. Checking the answer with FOIL is a normal part of factorisation: you will do it over and over again when you are solving factorisation problems.

The correct choice is: True.

Factorisation is the inverse of expanding something. When we factorise, we must break something apart into a product. For example, if you factorise the number $10$, you write $2 \cdot 5$. For a trinomial, it is common that the factors are binomials, like what is shown in this question.

To see if the binomials are the correct factors, we must expand the product to find out if they are equal to the trinomial $x^{2} - 7 x + 12$ or not. That means using FOIL:

These binomials are not the correct factors: we were supposed to get the trinomial $x^{2} - 7 x + 12$! However, the first and second terms are both wrong.

Notice that we checked the answer by expanding the binomials with FOIL and comparing the result to the trinomial in the question. Checking the answer with FOIL is a normal part of factorisation: you will do it over and over again when you are solving factorisation problems.

The correct choice is: False.

You have the expression $3 x^{2} + 11 x + 6$ and you must factorise it. Factorising an expression usually requires patience, but in this case it is worse than usual because the quadratic coefficient is not $1$: that adds to the complexity of the question.

Factorisation usually requires some trial and error, but you can usually use the quadratic term and the signs to start breaking down the factors. In this case:

• the $3 x^{2}$ term means that the factors must start with $3 x$ and $x$
• since all of the terms in the trinomial are positive, the factors must both have plus signs

That gives us:

(If you are not sure where these values came from, imagine starting FOIL with these two brackets and compare to the expression $3 x^{2} + 11 x + 6$.)

Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, they should have a product of $6$ so that we can get the last term in $3 x^{2} + 11 x + 6$.

For example, let's check what happens if we try the numbers $1$ and $6$:

Shucks - we did not get the trinomial we needed because the $x$-term is not correct. That means that the binomials $\left(3 x + 1\right)$ and $\left(x + 6\right)$ cannot be the correct factors.

You need to try another combination of numbers which have a product of $6$. Note that swapping the $1$ and $6$ will give a different answer because of the $3 x$. In other words, even though $\left(3 x + 1\right) \left(x + 6\right)$ is not correct, $\left(3 x + 6\right) \left(x + 1\right)$ might be correct.

The correct factors are $\left(3 x + 2\right)\left(x + 3\right)$; you can test this using FOIL, and you will get $3 x^{2} + 11 x + 6$.

The correct answer is: $\left(3 x + 2\right) \left(x + 3\right)$.

You have the expression $5 x^{2} + 6 x + 1$ and you must factorise it. Factorising an expression usually requires patience, but in this case it is worse than usual because the quadratic coefficient is not $1$: that adds to the complexity of the question.

Factorisation usually requires some trial and error, but you can usually use the quadratic term and the signs to start breaking down the factors. In this case:

• the $5 x^{2}$ term means that the factors must start with $5 x$ and $x$
• since all of the terms in the trinomial are positive, the factors must both have plus signs

That gives us:

(If you are not sure where these values came from, imagine starting FOIL with these two brackets and compare to the expression $5 x^{2} + 6 x + 1$.)

Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, they should have a product of $1$ so that we can get the last term in $5 x^{2} + 6 x + 1$.

You are in luck with this expression because the only way to get that $1$ is $1 \times 1$. (You still need to check the calculation to make sure that the trinomial is not prime.)

Nice! We got the trinomial we needed, so the binomials must be correct! It is quite nice that it worked out on the first try: it often takes more than one try.

The correct answer is: $\left(5 x + 1\right) \left(x + 1\right)$.

You have the expression $3 x^{2} + 10 x + 3$ and you must factorise it. Factorising an expression usually requires patience, but in this case it is worse than usual because the quadratic coefficient is not $1$: that adds to the complexity of the question.

Factorisation usually requires some trial and error, but you can usually use the quadratic term and the signs to start breaking down the factors. In this case:

• the $3 x^{2}$ term means that the factors must start with $3 x$ and $x$
• since all of the terms in the trinomial are positive, the factors must both have plus signs

That gives us:

(If you are not sure where these values came from, imagine starting FOIL with these two brackets and compare to the expression $3 x^{2} + 10 x + 3$.)

Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, they should have a product of $3$ so that we can get the last term in $3 x^{2} + 10 x + 3$.

For example, let's check what happens if we try the numbers $1$ and $3$:

Nice! We got the trinomial we needed, so the binomials must be correct! It is quite nice that it worked out on the first try: it often takes more than one try.

The correct answer is: $\left(3 x + 1\right) \left(x + 3\right)$.

Here are two things to keep in mind: Look at the constant term in the trinomial: it is negative, so the binomials must have opposite signs, like this: $\small \left(\underline{\quad} + \underline{\quad} \right) \left(\underline{\quad} - \underline{\quad} \right)$

For this question you must factorise the expression $2 a^{2} - 11 a - 6$. For the factors of $2 a^{2} - 11 a - 6$, the following things must be true:

• the coefficient of the quadratic term is a prime number, so the binomial factors must start with $2 a$ and $a$ (there are no other combinations that have a product of $2 a^{2}$)
• since the constant term is negative, the signs in the binomials must be opposites (one positive and one negative)

From that information, here is an intelligent starting point.

This is a good start, but keep in mind that it may prove to be wrong - it is not the only possibility; for example, the signs can be swapped, which would give a different answer.

Now try some numbers in the binomials to get the constant term of $6$. For example, see what happens if you try the numbers $6$ and $1$.

Unfortunately, those are not the correct factors. If there are any other possible combinations of numbers to put into the binomials, you must try them to see which combination works out. For example, you can try swapping the signs in the binomials: $\left(2 a - 6\right) \left(a + 1\right)$.

The correct answer is: $\left(2 a + 1\right) \left(a - 6\right)$.

Here are two things to keep in mind: Look at the constant term in the trinomial: it is negative, so the binomials must have opposite signs, like this: $\small \left(\underline{\quad} + \underline{\quad} \right) \left(\underline{\quad} - \underline{\quad} \right)$

For this question you must factorise the expression $6 x^{2} + 19 x - 7$. For the factors of $6 x^{2} + 19 x - 7$, the following things must be true:

• the coefficient of the quadratic term is a composite number, so there is more than one possibility for how to break it down in the binomial factors; the correct values might be $6 x$ and $x$, or they could be $3 x$ and $2 x$
• since the constant term is negative, the signs in the binomials must be opposites (one positive and one negative)

From that information, here is an intelligent starting point.

This is a good start, but keep in mind that it may prove to be wrong - it is not the only possibility; for example, the signs can be swapped, which would give a different answer.

Now try some numbers in the binomials to get the constant term of $7$. For example, see what happens if you try the numbers $7$ and $1$.

Great! We got the trinomial we needed, so the binomials must be correct.

The correct answer is: $\left(2 x + 7\right) \left(3 x - 1\right)$.

Here are two things to keep in mind: All of the terms are positive, which means that the binomials must have positive signs in them: $\small \left(\underline{\quad} + \underline{\quad} \right) \left(\underline{\quad} + \underline{\quad} \right)$

For this question you must factorise the expression $9 y^{2} + 17 y + 8$. For the factors of $9 y^{2} + 17 y + 8$, the following things must be true:

• the coefficient of the quadratic term is a composite number, so there is more than one possibility for how to break it down in the binomial factors; the correct values might be $9 y$ and $y$, or they could be $3 y$ and $3 y$
• since all of the terms in the trinomial are positive, the factors must both have plus signs

From that information, here is an intelligent starting point.

This is a good start, but keep in mind that it may prove to be wrong - it is not the only possibility because, for example, you could try different numbers in order to get the quadratic term, $9 y^{2}$.

Now try some numbers in the binomials to get the constant term of $8$. For example, see what happens if you try the numbers $8$ and $1$.

Great! We got the trinomial we needed, so the binomials must be correct.

The correct answer is: $\left(9 y + 8\right) \left(y + 1\right)$.

We have the equation $x^{2} - x - 42 = \left(x - 7\right) \left(x \, + \, \underline{\; ? \;}\right)$, which shows an unfinished factorisation question. We must complete the factorisation by filling in the missing number.

To factorise an expression means to write the expression as a product. Factorising the trinomial $x^{2} - x - 42$ leads to two binomials: $\left(x - 7\right) \left(x \, + \, \underline{\; ? \;}\right)$. We need to put a number into the blank space so that the binomials have a product equal to $x^{2} - x - 42$.

We must think about FOIL when we factorise into binomials: if we expand the binomials using FOIL we must get the trinomial $x^{2} - x - 42$ (because the product must be equal to that trinomial).

Imagine doing FOIL with these two brackets: the "Lasts" term must be equal to $-42$! What number should go into the blank space for that to work out?

Notice that the constant term in the trinomial is negative. That is why the signs in the binomials are opposites: negative multiplied by positive is negative! $-7 \times 6 = -42$ .

The complete factorised expression is $\left(x - 7\right) \left(x + 6\right)$, and the answer is: $6$.

We have the equation $x^{2} - 10 x + 25 = \left(x - 5\right) \left(x \, - \, \underline{\; ? \;}\right)$, which shows an unfinished factorisation question. We must complete the factorisation by filling in the missing number.

To factorise an expression means to write the expression as a product. Factorising the trinomial $x^{2} - 10 x + 25$ leads to two binomials: $\left(x - 5\right) \left(x \, - \, \underline{\; ? \;}\right)$. We need to put a number into the blank space so that the binomials have a product equal to $x^{2} - 10 x + 25$.

We must think about FOIL when we factorise into binomials: if we expand the binomials using FOIL we must get the trinomial $x^{2} - 10 x + 25$ (because the product must be equal to that trinomial).

Imagine doing FOIL with these two brackets: the "Lasts" term must be equal to $25$! What number should go into the blank space for that to work out?

Notice that the constant term in the trinomial is positive. That is why the signs in the binomials are the same (both negative): negative times negative is positive! $-5 \times -5 = 25$ .

The complete factorised expression is $\left(x - 5\right) \left(x - 5\right)$, and the answer is: $5$.

We have the equation $x^{2} + 9 x + 18 = \left(x \, + \, \underline{\; ? \;}\right) \left(x + 3\right)$, which shows an unfinished factorisation question. We must complete the factorisation by filling in the missing number.

To factorise an expression means to write the expression as a product. Factorising the trinomial $x^{2} + 9 x + 18$ leads to two binomials: $\left(x \, + \, \underline{\; ? \;}\right) \left(x + 3\right)$. We need to put a number into the blank space so that the binomials have a product equal to $x^{2} + 9 x + 18$.

We must think about FOIL when we factorise into binomials: if we expand the binomials using FOIL we must get the trinomial $x^{2} + 9 x + 18$ (because the product must be equal to that trinomial).

Imagine doing FOIL with these two brackets: the "Lasts" term must be equal to $18$! What number should go into the blank space for that to work out?

Notice that the constant term in the trinomial is positive. That is why the signs in the binomials are the same (both positive): positive times positive is positive! $6 \times 3 = 18$ .

The complete factorised expression is $\left(x + 6\right) \left(x + 3\right)$, and the answer is: $6$.

The first step in solving this problem is to factorise the binomial. This binomial is the difference of two squares. Once you have determined the two factors, apply the zero product rule by setting each factor equal to zero. You can then solve the two factors by isolating the variable, $a$.

$\therefore a= - \frac{10}{7}$ or $a= \frac{10}{7}$ .

The first step in solving this problem is to factorise the binomial. This binomial is the difference of two squares. Once you have determined the two factors, apply the zero product rule by setting each factor equal to zero. You can then solve the two factors by isolating the variable, $a$.

$\therefore a= - \frac{2}{3}$ or $a= \frac{2}{3}$ .

The first step in solving this problem is to factorise the binomial. This binomial is the difference of two squares. Once you have determined the two factors, apply the zero product rule by setting each factor equal to zero. You can then solve the two factors by isolating the variable, $x$.

$\therefore x= - \frac{5}{3}$ or $x= \frac{5}{3}$ .

The method for solving quadratic equations is factorisation. After you factorise, you will use the two binomials you get from factorisation to find the solutions to the equation.

Factorising the equation gives the following:

This means that one of these things must be true:

Therefore: $\; a = -7$ or $a = 6$.

The method for solving quadratic equations is factorisation. After you factorise, you will use the two binomials you get from factorisation to find the solutions to the equation.

Factorising the equation gives the following:

This means that one of these things must be true:

Therefore: $\; c = -7$ or $c = -3$.

The method for solving quadratic equations is factorisation. After you factorise, you will use the two binomials you get from factorisation to find the solutions to the equation.

Factorising the equation gives the following:

This means that one of these things must be true:

Therefore: $\; z = -5$ or $z = 8$.

Which two numbers will give $\text{225}$ when squared? Remember, when you square a negative number you get a positive number.

We already know how to solve linear equations using inverse operations. But, this approach does not always work for non-linear equations. This equation has a $p^{2}$ term, so it is not linear (it is quadratic).

When we solve equations, we are looking for the value of $p$ that will make the left hand side equal to the right hand side. We will solve non-linear equations using inspection, which means looking carefully. This question is asking "What number (or numbers), when you square it, gives $\text{225}$?".

Since $15 \times 15 = 225$, we know that $\text{15}$ is a solution.

However, this is not the only solution. Remember, when you square a negative number you get a positive number.

Therefore, $-\text{15}$ is also a solution. Either of the two answers will make the equation true, so we write

Which two numbers will give $\text{9}$ when squared? Remember, when you square a negative number you get a positive number.

We already know how to solve linear equations using inverse operations. But, this approach does not always work for non-linear equations. This equation has a $p^{2}$ term, so it is not linear (it is quadratic).

When we solve equations, we are looking for the value of $p$ that will make the left hand side equal to the right hand side. We will solve non-linear equations using inspection, which means looking carefully. This question is asking "What number (or numbers), when you square it, gives $\text{9}$?".

Since $3 \times 3 = 9$, we know that $\text{3}$ is a solution.

However, this is not the only solution. Remember, when you square a negative number you get a positive number.

Therefore, $-\text{3}$ is also a solution. Either of the two answers will make the equation true, so we write

Which two numbers will give $\text{144}$ when squared? Remember, when you square a negative number you get a positive number.

We already know how to solve linear equations using inverse operations. But, this approach does not always work for non-linear equations. This equation has a $b^{2}$ term, so it is not linear (it is quadratic).

When we solve equations, we are looking for the value of $b$ that will make the left hand side equal to the right hand side. We will solve non-linear equations using inspection, which means looking carefully. This question is asking "What number (or numbers), when you square it, gives $\text{144}$?".

Since $12 \times 12 = 144$, we know that $\text{12}$ is a solution.

However, this is not the only solution. Remember, when you square a negative number you get a positive number.

Therefore, $-\text{12}$ is also a solution. Either of the two answers will make the equation true, so we write

The equation looks confusing because it is not in standard form. In "standard form," the values of $a, b$ and $c$ are as follows: $\; 0 = ax^{2} + bx + c \;$ --or-- $\; ax^{2} + bx + c = 0\; \longrightarrow a$ is the coefficient of the quadratic term; $b$ is the coefficient of the linear term; $c$ is the value of the constant term.

• You should first apply the distributive law to both sides of the equation.
• Rearrange the equation into standard form by collecting all the terms on one side such that the value of $a$ is positive and the equation is equal to zero.
• Then the equation will look like a "normal" quadratic equation.
• Find the factors and then solve for $a$.

Therefore, the roots of the equation are $a = -9$ and $a = - \frac{5}{2}$ .

The equation looks confusing because it is not in standard form. In "standard form," the values of $a, b$ and $c$ are as follows: $\; 0 = ax^{2} + bx + c \;$ --or-- $\; ax^{2} + bx + c = 0\; \longrightarrow a$ is the coefficient of the quadratic term; $b$ is the coefficient of the linear term; $c$ is the value of the constant term.

• You should first apply the distributive law to both sides of the equation.
• Rearrange the equation into standard form by collecting all the terms on one side such that the value of $a$ is positive and the equation is equal to zero.
• Then the equation will look like a "normal" quadratic equation.
• Find the factors and then solve for $b$.

Therefore, the roots of the equation are $b = -6$ and $b = -6$ . This is a special case where both roots have the same solution, $b = -6$.

The equation looks confusing because it is not in standard form. In "standard form," the values of $a, b$ and $c$ are as follows: $\; 0 = ax^{2} + bx + c \;$ --or-- $\; ax^{2} + bx + c = 0\; \longrightarrow a$ is the coefficient of the quadratic term; $b$ is the coefficient of the linear term; $c$ is the value of the constant term.

• You should first apply the distributive law to both sides of the equation.
• Rearrange the equation into standard form by collecting all the terms on one side such that the value of $a$ is positive and the equation is equal to zero.
• Then the equation will look like a "normal" quadratic equation.
• Find the factors and then solve for $b$.

Therefore, the roots of the equation are $b = -2$ and $b = - \frac{6}{5}$ .

Start by arranging the equation in standard form; then factorise!

You must first arrange the equation in standard form (the zero will be very important below!):

Now factorise the trinomial on the left of the equation.

The equation above tells us that the binomials have a product of zero. That means that one of the binomials must be equal to zero:

The solutions to the equation are $b = -3$ or $b = 5$.

Start by arranging the equation in standard form; then factorise!

You must first arrange the equation in standard form (the zero will be very important below!):

Now factorise the trinomial on the left of the equation.

The equation above tells us that the binomials have a product of zero. That means that one of the binomials must be equal to zero:

The solutions to the equation are $m = -3$ or $m = -2$.

Start by arranging the equation in standard form; then factorise!

You must first arrange the equation in standard form (the zero will be very important below!):

Now factorise the trinomial on the left of the equation.

The equation above tells us that the binomials have a product of zero. That means that one of the binomials must be equal to zero:

The solutions to the equation are $y = -2$ or $y = 6$.

Standard form for a quadratic equation looks like this:

You just need to read the values of $a$, $b$, and $c$ from the equation.

The equation might look confusing because it is not in standard form. In "standard form," the values of $a$, $b$, and $c$ are as follows:

$a$ is the coefficient of the quadratic term; $b$ is the coefficient of the linear term; $c$ is the value of the constant term.

You can rearrange the equation by collecting all the terms on one side which will result in a $\text{0}$ on the other side of the equation. When you do this, make sure the quadratic coefficient is positive, and simplify the equation so that the powers of the terms are in descending order.

The quadratic equation is now in "standard form" and ready to be factorised. Solve for $y$ by making each factor equal to $0$, and isolate the variable.

The solutions for this equation are $y = -2$ and $y = - \dfrac{1}{4}$.

Standard form for a quadratic equation looks like this:

You just need to read the values of $a$, $b$, and $c$ from the equation.

The equation might look confusing because it is not in standard form. In "standard form," the values of $a$, $b$, and $c$ are as follows:

$a$ is the coefficient of the quadratic term; $b$ is the coefficient of the linear term; $c$ is the value of the constant term.

You can rearrange the equation by collecting all the terms on one side which will result in a $\text{0}$ on the other side of the equation. When you do this, make sure the quadratic coefficient is positive, and simplify the equation so that the powers of the terms are in descending order.

The quadratic equation is now in "standard form" and ready to be factorised. Solve for $x$ by making each factor equal to $0$, and isolate the variable.

The solutions for this equation are $x = - \dfrac{8}{3}$ and $x = -2$.

Standard form for a quadratic equation looks like this:

You just need to read the values of $a$, $b$, and $c$ from the equation.

The equation might look confusing because it is not in standard form. In "standard form," the values of $a$, $b$, and $c$ are as follows:

$a$ is the coefficient of the quadratic term; $b$ is the coefficient of the linear term; $c$ is the value of the constant term.

You can rearrange the equation by collecting all the terms on one side which will result in a $\text{0}$ on the other side of the equation. When you do this, make sure the quadratic coefficient is positive, and simplify the equation so that the powers of the terms are in descending order.

The quadratic equation is now in "standard form" and ready to be factorised. Solve for $z$ by making each factor equal to $0$, and isolate the variable.

The solutions for this equation are $z = -8$ and $z = - \dfrac{6}{5}$.

You need to solve this equation by factorisation, but there is a step you can do before you factorise which will help make the problem easier!

All of the terms in this equation have a factor of $\text{2}$. Divide the equation by $\text{2}$ to remove this common factor and simplify the equation. Factorisation is easier with smaller numbers!

Factorise the left hand side:

From the two binomials, you get two possibilities: either $c +3 = 0$ or $c +7 = 0$. From these two options, you get the answers: $\; c = -7$ or $c = -3$.

You need to solve this equation by factorisation, but there is a step you can do before you factorise which will help make the problem easier!

All of the terms in this equation have a factor of $\text{2}$. Divide the equation by $\text{2}$ to remove this common factor and simplify the equation. Factorisation is easier with smaller numbers!

Factorise the left hand side:

From the two binomials, you get two possibilities: either $a +4 = 0$ or $a +9 = 0$. From these two options, you get the answers: $\; a = -9$ or $a = -4$.

You need to solve this equation by factorisation, but there is a step you can do before you factorise which will help make the problem easier!

All of the terms in this equation have a factor of $\text{4}$. Divide the equation by $\text{4}$ to remove this common factor and simplify the equation. Factorisation is easier with smaller numbers!

Factorise the right hand side:

From the two binomials, you get two possibilities: either $x +2 = 0$ or $x +3 = 0$. From these two options, you get the answers: $\; x = -3$ or $x = -2$.

The first step in solving this problem is to factorise the trinomial. Once you have determined the two factors, apply the zero product rule by setting each factor equal to zero. You can then solve the two factors by isolating the variable, $y$.

$\therefore y = -9$ or $y = 2$.

The first step in solving this problem is to factorise the trinomial. Once you have determined the two factors, apply the zero product rule by setting each factor equal to zero. You can then solve the two factors by isolating the variable, $z$.

$\therefore z = -7$ or $z = 1$.

The first step in solving this problem is to factorise the trinomial. Once you have determined the two factors, apply the zero product rule by setting each factor equal to zero. You can then solve the two factors by isolating the variable, $a$.

$\therefore a = -5$ or $a = -3$.

The first step in solving this problem is to factorise the trinomial.

Once we have the two binomial factors, we use the zero product rule by setting each factor equal to zero. This leads to two separate equations. Then we can solve each of the equations by isolating the variable, $y$.

The correct answers are: $y = -7$ or $y = - \frac{1}{4}$.

The first step in solving this problem is to factorise the trinomial.

Once we have the two binomial factors, we use the zero product rule by setting each factor equal to zero. This leads to two separate equations. Then we can solve each of the equations by isolating the variable, $z$.

The correct answers are: $z = - \frac{9}{2}$ or $z = - \frac{6}{5}$.

The first step in solving this problem is to factorise the trinomial.

Once we have the two binomial factors, we use the zero product rule by setting each factor equal to zero. This leads to two separate equations. Then we can solve each of the equations by isolating the variable, $y$.

The correct answers are: $y = - \frac{7}{5}$ or $y = - \frac{1}{2}$.

Standard form for a quadratic equation is $ax^{2} + bx + c = 0$. You need to multiply the brackets and then simplify so that you can arrange the equation so that it matches standard form.

We must expand the brackets. Also, remember to bring the constant over from the other side of the equation because standard form for a quadratic equation requires a zero on one side of the equation:

Now simplify the expression on the left of the equation by collecting like terms.

The equation now matches the arrangement for standard form:$\;as^{2} + bs + c = 0$; so we are done!

The final answer is $10 s^{2} + 31 s + 19 = 0$

Standard form for a quadratic equation is $ax^{2} + bx + c = 0$. You need to multiply the brackets and then simplify so that you can arrange the equation so that it matches standard form.

We must expand the brackets. Also, remember to bring the constant over from the other side of the equation because standard form for a quadratic equation requires a zero on one side of the equation:

Now simplify the expression on the left of the equation by collecting like terms.

The equation now matches the arrangement for standard form:$\;at^{2} + bt + c = 0$; so we are done!

The final answer is $2 t^{2} + 13 t + 27 = 0$

Standard form for a quadratic equation is $ax^{2} + bx + c = 0$. You need to multiply the brackets and then simplify so that you can arrange the equation so that it matches standard form.

We must expand the brackets. Also, remember to bring the constant over from the other side of the equation because standard form for a quadratic equation requires a zero on one side of the equation:

Now simplify the expression on the left of the equation by collecting like terms.

The equation now matches the arrangement for standard form:$\;ap^{2} + bp + c = 0$; so we are done!

The final answer is $16 p^{2} + 56 p + 61 = 0$

The first step in solving this problem is to factorise the binomial. This binomial is the the difference of two squares. When you have the two factors you set them equal to zero, applying the zero product rule. You then solve the two factors by isolating the variable, $y$.

$\therefore y = -3$ or $y= 3$

The first step in solving this problem is to factorise the binomial. This binomial is the the difference of two squares. When you have the two factors you set them equal to zero, applying the zero product rule. You then solve the two factors by isolating the variable, $y$.

$\therefore y = -6$ or $y= 6$

The first step in solving this problem is to factorise the binomial. This binomial is the the difference of two squares. When you have the two factors you set them equal to zero, applying the zero product rule. You then solve the two factors by isolating the variable, $a$.

$\therefore a = -5$ or $a= 5$

The first step in solving this problem is to factorise the trinomial. When you have the two factors you set them equal to zero, applying the zero product rule. You then solve the two factors by isolating the variable, $y$.

$\therefore y = -4$ or $y=-4$. This is a special case where both the roots have the same answer, $y = -4$.

The first step in solving this problem is to factorise the trinomial. When you have the two factors you set them equal to zero, applying the zero product rule. You then solve the two factors by isolating the variable, $b$.

$\therefore b = -5$ or $b=-1$.

The first step in solving this problem is to factorise the trinomial. When you have the two factors you set them equal to zero, applying the zero product rule. You then solve the two factors by isolating the variable, $b$.

$\therefore b = -6$ or $b=-2$.

The first step in solving this problem is to factorise the trinomial. Once we have determined the two factors, we use the zero product rule by setting each factor equal to zero. We can then solve the two factors by isolating the variable, $z$.

The solutions to the equation are $z = - \frac{9}{2}$ and $z = -1$.

The first step in solving this problem is to factorise the trinomial. Once we have determined the two factors, we use the zero product rule by setting each factor equal to zero. We can then solve the two factors by isolating the variable, $y$.

The solutions to the equation are $y = -8$ and $y = \frac{5}{4}$.

The first step in solving this problem is to factorise the trinomial. Once we have determined the two factors, we use the zero product rule by setting each factor equal to zero. We can then solve the two factors by isolating the variable, $x$.

The solutions to the equation are $x = -5$ and $x = \frac{3}{4}$.

Which two numbers will give $\text{144}$ when squared? Remember, when you multiply a negative number by itself you get a positive number.

We already know how to solve linear equations using inverse operations. But, this approach does not always work for non-linear equations. This equation has a $b^{2}$ term, so it is not linear (it is quadratic). We use inspection to solve non-linear equations.

We will start by solving this equation for the whole bracket:

To answer this question we ask "What number (or numbers), when you square it, gives you $\text{144}$?".

Since $12 \times 12 = 144$, we know that $\text{12}$ is a value for ${\color{red}{b - 3}}$.

However, this is not the only number that gives $\text{144}$ when multiplied by itself. Remember, when you square a negative number you get a positive number:

Therefore, ${\color{red}{b - 3}}$ must be equal to ${\color{red}{12}}$ or ${\color{red}{-12}}$.

Mathematically we write this as:

Now we have created two linear equations that can each be solved separately for $b$, using inverse operations.

or

Which two numbers will give $\text{64}$ when squared? Remember, when you multiply a negative number by itself you get a positive number.

We already know how to solve linear equations using inverse operations. But, this approach does not always work for non-linear equations. This equation has a $a^{2}$ term, so it is not linear (it is quadratic). We use inspection to solve non-linear equations.

We will start by solving this equation for the whole bracket:

To answer this question we ask "What number (or numbers), when you square it, gives you $\text{64}$?".

Since $8 \times 8 = 64$, we know that $\text{8}$ is a value for ${\color{red}{a - 1}}$.

However, this is not the only number that gives $\text{64}$ when multiplied by itself. Remember, when you square a negative number you get a positive number:

Therefore, ${\color{red}{a - 1}}$ must be equal to ${\color{red}{8}}$ or ${\color{red}{-8}}$.

Mathematically we write this as:

Now we have created two linear equations that can each be solved separately for $a$, using inverse operations.

or

Which two numbers will give $\text{81}$ when squared? Remember, when you multiply a negative number by itself you get a positive number.

We already know how to solve linear equations using inverse operations. But, this approach does not always work for non-linear equations. This equation has a $y^{2}$ term, so it is not linear (it is quadratic). We use inspection to solve non-linear equations.

We will start by solving this equation for the whole bracket:

To answer this question we ask "What number (or numbers), when you square it, gives you $\text{81}$?".

Since $9 \times 9 = 81$, we know that $\text{9}$ is a value for ${\color{red}{y + 1}}$.

However, this is not the only number that gives $\text{81}$ when multiplied by itself. Remember, when you square a negative number you get a positive number:

Therefore, ${\color{red}{y + 1}}$ must be equal to ${\color{red}{9}}$ or ${\color{red}{-9}}$.

Mathematically we write this as:

Now we have created two linear equations that can each be solved separately for $y$, using inverse operations.

or

You know that $x = 6$ is a root of the equation; that means that when $x = 6, y = 0$. Substitute these numbers into the equation, and then tidy up as much as you can:

Now just solve for $k$.

The value of $k$ is $-\text{15}$.

Now you can move on to the second part of the question: finding the other root of the equation. First, you can now write the full equation, becuase you know that $k = -15$:

We want to find a root, which means (as always!) $y = 0$. To solve for the root, you need to factorise - and it is very helpful to remember that you already know one of the roots ($x = 6$) !

By comparing these two lines you can see that the other binomial must contain a $\text{2}$ and a $-\text{3}$...

Now just solve for $x$ using the first binomial to get the other root.

The second root is $x = \frac{3}{2}$.

You know that $x = -1$ is a root of the equation; that means that when $x = -1, y = 0$. Substitute these numbers into the equation, and then tidy up as much as you can:

Now just solve for $k$.

The value of $k$ is $-\text{10}$.

Now you can move on to the second part of the question: finding the other root of the equation. First, you can now write the full equation, becuase you know that $k = -10$:

We want to find a root, which means (as always!) $y = 0$. To solve for the root, you need to factorise - and it is very helpful to remember that you already know one of the roots ($x = -1$) !

By comparing these two lines you can see that the other binomial must contain a $\text{3}$ and a $-\text{10}$...

Now just solve for $x$ using the first binomial to get the other root.

The second root is $x = \frac{10}{3}$.

You know that $x = -3$ is a root of the equation; that means that when $x = -3, y = 0$. Substitute these numbers into the equation, and then tidy up as much as you can:

Now just solve for $m$.

The value of $m$ is $\text{2}$.

Now you can move on to the second part of the question: finding the other root of the equation. First, you can now write the full equation, becuase you know that $m = 2$:

We want to find a root, which means (as always!) $y = 0$. To solve for the root, you need to factorise - and it is very helpful to remember that you already know one of the roots ($x = -3$) !

By comparing these two lines you can see that the other binomial must contain a $\text{2}$ and a $-\text{5}$...

Now just solve for $x$ using the first binomial to get the other root.

The second root is $x = \frac{5}{2}$.